Question

Prove that cot $(\frac{\pi }{4}-2co{t}^{-1}3)=7$

Answer 1

We have to prove, cot$(\frac{\pi }{4}-2co{t}^{-1}3)=7$

$\Rightarrow (\frac{\pi }{4}-2co{t}^{-1}3)=co{t}^{-1}7\Rightarrow \left(2co{t}^{-1}3\right)=\frac{\pi }{4}-co{t}^{-1}7\Rightarrow 2ta{n}^{-1}\frac{1}{3}=\frac{\pi }{4}-ta{n}^{-1}\frac{1}{7}\Rightarrow 2ta{n}^{-1}\frac{1}{3}+ta{n}^{-1}\frac{1}{7}=\frac{\pi }{4}\Rightarrow ta{n}^{-1}\frac{2/3}{1-(1/3{)}^2}+ta{n}^{-1}\frac{1}{7}=\frac{\pi }{4}\Rightarrow ta{n}^{-1}\frac{2/3}{8/9}+ta{n}^{-1}\frac{1}{7}=\frac{\pi }{4}\Rightarrow ta{n}^{-1}\frac{3}{4}+ta{n}^{-1}\frac{1}{7}=\frac{\pi }{4}\Rightarrow ta{n}^{-1}\frac{\frac{3}{4}+\frac{1}{7}}{1-\frac{3}{4}.\frac{1}{7}}=\frac{\pi }{4}\Rightarrow ta{n}^{-1}\frac{(21+4)/28}{(28-3)/28}=\frac{\pi }{4}\Rightarrow ta{n}^{-1}\frac{25}{25}=\frac{\pi }{4}\Rightarrow 1=tan\frac{\pi }{4}\Rightarrow 1=1\Rightarrow LHS=RHSHenceproved.$